Câu hỏi 23

Đặt \({V_{ABCD.A'B'C'D'}} = V\).

Ta có: \({V_{ABCD.A'B'C'D'}} = {V_{ACB'D'}} + {V_{B'.ABC}} + {V_{D'.ACD}} + {V_{A.A'B'D'}} + {V_{C.B'C'D'}}\).

Ta có: \({V_{B'.ABC}} = \dfrac{1}{3}d\left( {B';\left( {ABC} \right)} \right).{S_{\Delta ABC}}\)\( = \dfrac{1}{3}d\left( {B';\left( {ABCD} \right)} \right).\dfrac{1}{2}{S_{ABCD}} = \dfrac{1}{6}V\).

Chứng minh tương tự ta có:  \({V_{D'.ACD}} = {V_{A.A'B'D'}} = {V_{C.B'C'D'}} = \dfrac{1}{6}V\).

 Khi đó ta có: \(V = {V_{ACB'D'}} + 4.\dfrac{1}{6}V \Leftrightarrow {V_{ACB'D'}} = \dfrac{V}{3}\).

Vậy \(\dfrac{{{V_{ACB'D'}}}}{{{V_{ABCD.A'B'C'D'}}}} = \dfrac{V}{3}:V = \dfrac{1}{3}.\)

Chọn D.